An electrolyte solution has an average current density of $1$ ampere per square decimeter $\left( \dfrac{\text{A}}{\text{dm}^2}\right)$. What is the current density of the solution in $\dfrac{\text{A}}{\text{m}^2}$ ?
Solution: We will convert $1\, \dfrac{\text{A}}{\text{dm}^2}$ to a rate in $\dfrac{\text{A}}{\text{m}^2}$ using the following conversion rate: There are $10\text{ dm}$ per $1\text{ m}$. Therefore, there are $(10\text{ dm})^2=100\text{ dm}^2$ per $1\text{ m}^2$. $\begin{aligned} &\phantom{=} \dfrac{1\text{ A}}{1\text{ dm}^2} \cdot \dfrac{100\text{ dm}^2}{1 \text{ m}^2} \\\\ &=\dfrac{1\cdot 100\cdot\text{A} \cdot \cancel{\text{dm}^2} }{1 \cdot 1 \cdot \cancel{\text{dm}^2} \cdot \text{m}^2} \\\\ &=\dfrac{100}{1}\,\dfrac{\text{A}}{\text{m}^2} \\\\ &=100\,\dfrac{\text{A}}{\text{m}^2} \end{aligned}$ In conclusion, the current density of the solution in $\dfrac{\text{A}}{\text{m}^2}$ is: $100\,\dfrac{\text{A}}{\text{m}^2}$